A) (1, 1, 3)
B) (1, 1, 0)
C) \[\left( \frac{1}{2},\,\,2,\,\,0 \right)\]
D) \[\left( \frac{1}{2},\,\,2,\,\,3 \right)\]
Correct Answer: A
Solution :
\[{{y}^{2}}=8ax\] Normal \[y=mx-4\text{ }am-2a{{m}^{3}}\] ..... (1) \[{{y}^{2}}=4b\left( x-c \right)\] Normal \[y=m\left( x-c \right)-2\text{ }bm-2bm{{}^{3}}\] \[y=mx-mc-2\text{ }bm-2bm{{}^{3}}\] ..... (2) For common normal; compare eq. (1) & eq. (2) \[\frac{1}{1}=\,\frac{m}{m'}\,=\,\frac{4\,am+2\,a{{m}^{3}}}{m'c\,+\,2bm'+bm{{'}^{3}}}\] \[m=m';{m}'c+2b{m}'+b{{{m}'}^{3}}\,=\,4am+2a{{m}^{3}}\] \[\,m\left( {{m}^{2}}\left( 2a-b \right)+4a-2b-c \right)=0\]| Case (I) = 0 | Case (II) \[m\ne 0\] |
| x-axis is common normal of both curve | \[{{m}^{2}}\,=\,\frac{2b+c-4a}{2a-b}\,\,>0\] \[=\,\,\frac{c}{2a-b}\,\,-\,\,2\,\,>\,\,0\] |
| All option are correct (a), (b), (c), (d) | Only option (a), (1, 1, 3) satisfy it |
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